# Involves finding the intervals where a function f is increasing and decreasing. Given the function f(x)={(x+7,if x lt-3,,),(|x+1|,if -3 leq x<1,,),(5-2x,if x ge 1,,):}

Increasing and decreasing piecewise function on an interval
I'm working on a problem that involves finding the intervals where a function f is increasing and decreasing. Given the function
I worked out that f is increasing on $\left(-\mathrm{\infty },-3\right)$ and $\left[-1,1\right)$, and f is decreasing on $\left[-3,-1\right],\left[1,\mathrm{\infty }\right)$.
However, the solution in the book claims that f is increasing on [-1,1], rather than [-1,1) like I worked out. I am having trouble understanding why the book claims that this is the case.
I was under the impression that f must be differentiable on the interior of an interval I and continuous on all of I in order to make any statements about increasing/decreasing behavior on the closed interval I. I'm not sure if I am overlooking something, but it seems that f is not continuous on [-1,1]. I would appreciate any clarification.
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Alvin Preston
Step 1
It is true that if you have a differentiable function on an interval, then it is increasing if and only if its derivative is non-negative. However, increasing functions need not be differentiable according to their definition:
A function $f:\mathbb{R}\to \mathbb{R}$ is increasing on a collection S if and only if:
For any $x,y\in S$ such that $x\le y$:
$f\left(x\right)\le f\left(y\right)$
Step 2
Note that this definition is incompatible with the one that lulu proposed in a comment, but I believe this is the typical definition.
###### Did you like this example?
Deanna Gregory
Explanation:
You have found that f is increasing on $\left[-1,1\right)$
Now $\underset{x\to {1}^{-}}{lim}f\left(x\right)=|1+1|=2<3=f\left(1\right)$.
Thus f is increasing on $\left[-1,1\right]$